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During a decay of 952 gm $_{92} U^{238}$ to $_{82} {Pb}^{206}$ , total $18 x 10^{24}$ $α -$ particles are emitted, number of half lives completed in this decay are

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a

6

b

3

c

4

d

5

answer is B.

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Detailed Solution

$\begin{array}{l} _{92} U^{238} ⟶_{82} {Pb}^{206} + 8_{2} {He}^{4} + 6_{- 1} β^{0} \\ (a - x \\ (a = \frac{952}{238} = 4 moles) \\ 8 x = \frac{18 \times 10^{24}}{6 \times 10^{23}} = 30 \Rightarrow x = \frac{30}{8} \\ λ \times t = \ln (\frac{a}{a - x}) \\ \frac{\ln 2}{t_{1 / 2}} \times n \times t_{1 / 2} = \ln (\frac{4}{4 - \frac{30}{8}}) \Rightarrow n = 4 \end{array}$

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